Cramer's rule
You are encouraged to solve this task according to the task description, using any language you may know.
In linear algebra, Cramer's rule is an explicit formula for the solution of a system of linear equations with as many equations as unknowns, valid whenever the system has a unique solution. It expresses the solution in terms of the determinants of the (square) coefficient matrix and of matrices obtained from it by replacing one column by the vector of right hand sides of the equations.
Given
which in matrix format is
Then the values of and can be found as follows:
- Task
Given the following system of equations:
solve for , , and , using Cramer's rule.
EchoLisp
<lang scheme> (lib 'matrix) (string-delimiter "") (define (cramer A B (X)) ;; --> vector X (define ∆ (determinant A)) (for/vector [(i (matrix-col-num A))] (set! X (matrix-set-col! (array-copy A) i B)) (// (determinant X) ∆)))
(define (task) (define A (list->array
'( 2 -1 5 1 3 2 2 -6 1 3 3 -1 5 -2 -3 3) 4 4))
(define B #(-3 -32 -47 49)) (writeln "Solving A * X = B") (array-print A) (writeln "B = " B) (writeln "X = " (cramer A B))) </lang>
- Output:
(task) Solving A * X = B 2 -1 5 1 3 2 2 -6 1 3 3 -1 5 -2 -3 3 B = #( -3 -32 -47 49) X = #( 2 -12 -4 1)
Fortran
In Numerical Methods That Work (Usually), in the section What not to compute, F. S. Acton remarks "...perhaps we should be glad he didn't resort to Cramer's rule (still taught as the practical method in some high schools) and solve his equations as the ratios of determinants - a process that requires labor proportional to N! if done in the schoolboy manner. The contrast with N3 can be startling!" And further on, "Having hinted darkly at my computational fundamentalism, it is probably time to commit to a public heresy by denouncing recursive calculations. I have never seen a numerical problem arising from the physical world that was best calculated by a recursive subroutine..."
Since this problem requires use of Cramer's rule, one might as well be hung for a sheep instead of a lamb, so the traditions of Old Fortran and heavy computation will be ignored and the fearsome RECURSIVE specification employed so that the determinants will be calculated recursively, all the way down to N = 1 even though the N = 2 case is easy. This requires F90 and later. Similarly, the MODULE protocol will be employed, even though there is no significant context to share. The alternative method for calculating a determinant involves generating permutations, a tiresome process.
Array passing via the modern arrangements of F90 is a source of novel difficulty to set against the slight convenience of not having to pass an additional parameter, N. Explicitly, at least. There are "secret" additional parameters when an array is being passed in the modern way, which are referred to by the new SIZE function. Anyway, for an order N square matrix, the array must be declared A(N,N), and specifically not something like A(100,100) with usage only of elements up to N = 7, say, because the locations in storage of elements in use would be quite different from those used by an array declared A(7,7). This means that the array must be re-declared for each different size usage, a tiresome and error-inviting task. One-dimensional arrays do not have this problem, but they do have to be "long enough" so B and X might as well be included. This also means that the auxiliary matrices needed within the routines have to be made the right size, and fortunately they can be declared in a way that requests this without the blather of ALLOCATE, this being a protocol introduced by Algol in the 1960s. Unfortunately, there is no scheme such as in pl/i to declare AUX "like" A, so some grotesquery results. And in the case of function DET which needs an array of order N - 1, when its recursion bottoms out with N = 1 it will have declared MINOR(0,0), a rather odd situation that fortunately evokes no complaint, and a test run in which its "value" was written out by WRITE (6,*) MINOR produced a blank line: no complaint there either, presumably because zero elements were being sent forth and so there was no improper access of ... nothing.
With matrices, there is a problem all the way from the start in 1958. Everyone agrees that a matrix should be indexed as A(row,column) and that when written out, rows should run down the page and columns across. This is unexceptional and the F90 function MATMUL follows this usage. However, Fortran has always stored its array elements in what is called "column major" order, which is to say that element A(1,1) is followed by element A(2,1) in storage, not A(1,2). Thus, if an array is written (or read) by something like WRITE (6,*) A, consecutive elements, written along a line, will be A(1,1), A(2,1), A(3,1), ... So, subroutine SHOWMATRIX is employed to write the matrix out in the desired form, and to read the values into the array, an explicit loop is used to place them where expected rather than just READ(INF,*) A
Similarly, if instead a DATA statement were used to initialise the array for the example problem, and it looked something like <lang Fortran> DATA A/2, -1, 5, 1
1 3, 2, 2, -6
2 1, 3, 3, -1
3 5, -2, -3, 3/</lang> (ignoring integer/floating-point issues) thus corresponding to the layout of the example problem, there would need to be a statement A = TRANSPOSE(A) to obtain the required order.
I have never seen an explanation of why this choice was made for Fortran.<lang Fortran> MODULE BAD IDEA !Employ Cramer's rule for solving A.x = b...
INTEGER MSG !Might as well be here.
CONTAINS !The details.
SUBROUTINE SHOWMATRIX(A) !With nice vertical bars.
DOUBLE PRECISION A(:,:) !The matrix.
INTEGER R,N !Assistants.
N = SIZE(A, DIM = 1) !Instead of passing an explicit parameter.
DO R = 1,N !Work down the rows.
WRITE (MSG,1) A(R,:) !Fling forth a row at a go.
1 FORMAT ("|",<N>F12.3,"|") !Bounded by bars.
END DO !On to the next row.
END SUBROUTINE SHOWMATRIX !Furrytran's default order is the transpose.
RECURSIVE DOUBLE PRECISION FUNCTION DET(A) !Determine the determinant.
DOUBLE PRECISION A(:,:) !The square matrix, order N.
DOUBLE PRECISION MINOR(SIZE(A,DIM=1) - 1,SIZE(A,DIM=1) - 1) !Order N - 1.
DOUBLE PRECISION D !A waystation.
INTEGER C,N !Steppers.
N = SIZE(A, DIM = 1) !Suplied via secret parameters.
IF (N .LE. 0) THEN !This really ought not happen.
STOP "DET: null array!" !But I'm endlessly suspicious.
ELSE IF (N .NE. SIZE(A, DIM = 2)) THEN !And I'd rather have a decent message
STOP "DET: not a square array!" !In place of a crashed run.
ELSE IF (N .EQ. 1) THEN !Alright, now get on with it.
DET = A(1,1) !This is really easy.
ELSE !But otherwise,
D = 0 !Here we go.
DO C = 1,N !Step along the columns of the first row.
CALL FILLMINOR(C) !Produce the auxiliary array for each column.
IF (MOD(C,2) .EQ. 0) THEN !Odd or even case?
D = D - A(1,C)*DET(MINOR) !Even: subtract.
ELSE !Otherwise,
D = D + A(1,C)*DET(MINOR) !Odd: add.
END IF !So much for that term.
END DO !On to the next.
DET = D !Declare the result.
END IF !So much for that.
CONTAINS !An assistant.
SUBROUTINE FILLMINOR(CC) !Corresponding to A(1,CC).
INTEGER CC !The column being omitted.
INTEGER R !A stepper.
DO R = 2,N !Ignoring the first row,
MINOR(R - 1,1:CC - 1) = A(R,1:CC - 1) !Copy columns 1 to CC - 1. There may be none.
MINOR(R - 1,CC:) = A(R,CC + 1:) !And from CC + 1 to N. There may be none.
END DO !On to the next row.
END SUBROUTINE FILLMINOR !Divide and conquer.
END FUNCTION DET !Rather than mess with permutations.
SUBROUTINE CRAMER(A,X,B) !Solve A.x= b, where A is a matrix...
Careful! The array must be A(N,N), and not say A(100,100) of which only up to N = 6 are in use.
DOUBLE PRECISION A(:,:) !A square matrix. I hope.
DOUBLE PRECISION X(:),B(:) !Column matrices look rather like 1-D arrays.
DOUBLE PRECISION AUX(SIZE(A,DIM=1),SIZE(A,DIM=1)) !Can't say "LIKE A", as in pl/i, alas.
DOUBLE PRECISION D !To be calculated once.
INTEGER N !The order of the square matrix. I hope.
INTEGER C !A stepper.
N = SIZE(A, DIM = 1) !Alright, what's the order of battle?
D = DET(A) !Once only.
IF (D.EQ.0) STOP "Cramer: zero determinant!" !Surely, this won't happen...
AUX = A !Prepare the assistant.
DO C = 1,N !Step across the columns.
IF (C.GT.1) AUX(1:N,C - 1) = A(1:N,C - 1) !Repair previous damage.
AUX(1:N,C) = B(1:N) !Place the current damage.
X(C) = DET(AUX)/D !The result!
END DO !On to the next column.
END SUBROUTINE CRAMER !This looks really easy!
END MODULE BAD IDEA !But actually, it is a bad idea for N > 2.
PROGRAM TEST !Try it and see.
USE BAD IDEA !Just so.
DOUBLE PRECISION, ALLOCATABLE ::A(:,:), B(:), X(:) !Custom work areas.
INTEGER N,R !Assistants..
INTEGER INF !An I/O unit.
MSG = 6 !Output.
INF = 10 !For reading test data.
OPEN (INF,FILE="Test.dat",STATUS="OLD",ACTION="READ") !As in this file..
Chew into the next problem.
10 IF (ALLOCATED(A)) DEALLOCATE(A) !First,
IF (ALLOCATED(B)) DEALLOCATE(B) !Get rid of
IF (ALLOCATED(X)) DEALLOCATE(X) !The hired help.
READ (INF,*,END = 100) N !Since there is a new order.
IF (N.LE.0) GO TO 100 !Perhaps a final order.
WRITE (MSG,11) N !Othewise, announce prior to acting.
11 FORMAT ("Order ",I0," matrix A, as follows...") !In case something goes wrong.
ALLOCATE(A(N,N)) !For instance,
ALLOCATE(B(N)) !Out of memory.
ALLOCATE(X(N)) !But otherwise, a tailored fit.
DO R = 1,N !Now read in the values for the matrix.
READ(INF,*,END=667,ERR=665) A(R,:),B(R) !One row of A at a go, followed by B's value.
END DO !In free format.
CALL SHOWMATRIX(A) !Show what we have managed to obtain.
WRITE (MSG,12) "In the scheme A.x = b, b = ",B !In case something goes wrong.
12 FORMAT (A,<N>F12.6) !How many would be too many?
CALL CRAMER(A,X,B) !The deed!
WRITE (MSG,12) " Via Cramer's rule, x = ",X !The result!
GO TO 10 !And try for another test problem.
Complaints.
665 WRITE (MSG,666) "Format error",R !I know where I came from.
666 FORMAT (A," while reading row ",I0,"!") !So I can refer to R.
GO TO 100 !So much for that.
667 WRITE (MSG,666) "End-of-file", R !Some hint as to where.
Closedown.
100 WRITE (6,*) "That was interesting." !Quite.
END !Open files are closed, allocated memory is released. </lang>
Oddly, the Compaq Visual Fortran F90/95 compiler is confused by the usage "BAD IDEA" instead of "BADIDEA" - spaces are not normally relevant in Fortran source. Anyway, file Test.dat has been filled with the numbers of the example, as follows:
4 /The order, for A.x = b. 2 -1 5 1, -3 /First row of A, b 3 2 2 -6, -32 /Second row... 1 3 3 -1, -47 third row. 5 -2 -3 3, 49 /Last row.
Fortran's free-form allows a comma, a tab, and spaces between numbers, and regards the / as starting a comment, but, because each row is read separately, once the required five (N + 1) values have been read, no further scanning of the line takes place and the next READ statement will start with a new line of input. So the / isn't needed for the third row, as shown. Omitted values lead to confusion as the input process would read additional lines to fill the required count and everything gets out of step. Echoing input very soon after it is obtained is helpful in making sense of such mistakes.
For more practical use it would probably be better to constrain the freedom somewhat, perhaps requiring that all the N + 1 values for a row appear on one input record. In such a case, the record could first be read into a text variable (from which the data would be read) so that if a problem arises the text could be printed as a part of the error message. But, this requires guessing a suitably large length for the text variable so as to accommodate the longest possible input line.
Output:
Order 4 matrix A, as follows...
| 2.000 -1.000 5.000 1.000|
| 3.000 2.000 2.000 -6.000|
| 1.000 3.000 3.000 -1.000|
| 5.000 -2.000 -3.000 3.000|
In the scheme A.x = b, b = -3.000000 -32.000000 -47.000000 49.000000
Via Cramer's rule, x = 2.000000 -12.000000 -4.000000 1.000000
That was interesting.
And at this point I suddenly noticed that the habits of Old Fortran are not so easily suppressed: all calculations are done with double precision. Curiously enough, for the specific example data, the same results are obtained if all variables are integer.
J
Implementation:
<lang J>cramer=:4 :0
A=. x [ b=. y det=. -/ .* A %~&det (i.#A) b"_`[`]}&.|:"0 2 A
)</lang>
Task data:
<lang J>A=: _&".;._2]t=: 0 :0
2 -1 5 1 3 2 2 -6 1 3 3 -1 5 -2 -3 3
)
b=: _3 _32 _47 49</lang>
Task example:
<lang J> A cramer b 2 _12 _4 1</lang>
Perl
<lang perl>use Math::Matrix;
sub cramers_rule {
my ($A, $terms) = @_;
my @solutions;
my $det = $A->determinant;
foreach my $i (0 .. $#{$A}) {
my $Ai = $A->clone;
foreach my $j (0 .. $#{$terms}) {
$Ai->[$j][$i] = $terms->[$j];
}
push @solutions, $Ai->determinant / $det;
}
@solutions;
}
my $matrix = Math::Matrix->new(
[2, -1, 5, 1], [3, 2, 2, -6], [1, 3, 3, -1], [5, -2, -3, 3],
);
my $free_terms = [-3, -32, -47, 49]; my ($w, $x, $y, $z) = cramers_rule($matrix, $free_terms);
print "w = $w\n"; print "x = $x\n"; print "y = $y\n"; print "z = $z\n";</lang>
- Output:
w = 2 x = -12 y = -4 z = 1
Perl 6
<lang perl6>sub det(@matrix) {
my @a = @matrix.map: { [|$_] };
my $sign = +1;
my $pivot = 1;
for ^@a -> $k {
my @r = ($k+1 .. @a.end);
my $previous-pivot = $pivot;
if 0 == ($pivot = @a[$k][$k]) {
(my $s = @r.first: { @a[$_][$k] != 0 }) // return 0;
(@a[$s],@a[$k]) = (@a[$k], @a[$s]);
my $pivot = @a[$k][$k];
$sign = -$sign;
}
for @r X @r -> ($i, $j) {
((@a[$i][$j] *= $pivot) -= @a[$i][$k]*@a[$k][$j]) /= $previous-pivot;
}
}
$sign * $pivot
}
sub cramers_rule(@A, @terms) {
gather for ^@A -> $i {
my @Ai = @A.map: { [|$_] };
for ^@terms -> $j {
@Ai[$j][$i] = @terms[$j];
}
take det(@Ai);
} »/» det(@A);
}
my @matrix = (
[2, -1, 5, 1], [3, 2, 2, -6], [1, 3, 3, -1], [5, -2, -3, 3],
);
my @free_terms = (-3, -32, -47, 49); my ($w, $x, $y, $z) = |cramers_rule(@matrix, @free_terms);
say "w = $w"; say "x = $x"; say "y = $y"; say "z = $z";</lang>
- Output:
w = 2 x = -12 y = -4 z = 1
Ruby
<lang ruby>require 'matrix'
def cramers_rule(a, terms)
solutions = []
det = a.determinant
size = a.row_size;
size.times do |i|
cols = []
(0..i-1).each { |j|
cols << a.column(j)
}
cols << terms
(i+1...size).each { |j|
cols << a.column(j)
}
ai = Matrix.columns(cols)
solutions << ai.determinant / det
end
solutions
end
matrix = Matrix[
[2, -1, 5, 1], [3, 2, 2, -6], [1, 3, 3, -1], [5, -2, -3, 3],
]
vector = [-3, -32, -47, 49] puts cramers_rule(matrix, vector)</lang>
- Output:
2 -12 -4 1
Sidef
<lang ruby>func det(a) {
a = a.map{.map{_}}
var sign = +1
var pivot = 1
for k in ^a {
var r = (k+1 .. a.end)
var previous_pivot = pivot
if ((pivot = a[k][k])==0) {
a.swap(r.first_by { |i|
a[i][k] != 0
} \\ (return 0), k)
pivot = a[k][k]
sign = -sign
}
for i,j in (r ~X r) {
((a[i][j] *= pivot)
-= a[i][k]*a[k][j]
) /= previous_pivot
}
}
sign * pivot
}
func cramers_rule(A, terms) {
gather {
for i in ^A {
var Ai = A.map{.map{_}}
for j in ^terms {
Ai[j][i] = terms[j]
}
take(det(Ai))
}
} »/» det(A)
}
var matrix = [
[2, -1, 5, 1], [3, 2, 2, -6], [1, 3, 3, -1], [5, -2, -3, 3],
]
var free_terms = [-3, -32, -47, 49] var (w, x, y, z) = cramers_rule(matrix, free_terms)...
say "w = #{w}" say "x = #{x}" say "y = #{y}" say "z = #{z}"</lang>
- Output:
w = 2 x = -12 y = -4 z = 1